Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 216 / 2 / a / Solution

For the planar uniform distribution with a Lebesgue measure density, assume that the compact convex set $\mathcal{C}$ has nonempty interior, equivalently positive area. This assumption is missing from the wording: a line segment or singleton would not have such a planar uniform density.

The hit-and-run sampler at a current point $x$ chooses $u=(\cos \Theta ,\sin \Theta )$ with $\Theta$ uniform on $[0,2\pi )$, finds the intervaland samples $T$ uniformly on that interval, setting $x^{\prime }=x+Tu$. Convexity makes the intersection an interval and compactness makes it bounded.

If $x$ is in the interior, every direction has positive chord length, and the next point is in the interior almost surely. For a definition valid at every boundary point as well, redraw the direction whenever the chord has zero length. We use this convention in the following parts. Positive-length directions have positive angular probability at every $x$, since an interior ball is visible from $x$, so the redraw terminates almost surely.

The Markov kernel has the uniform distribution on $\mathcal{C}$ as its invariant law, as follows from the density symmetry proved next. Repeating the step therefore produces approximate uniform samples after convergence. The boundary convention matters: if singleton chords were instead treated as holding moves, a square's corner would have a positive atom at its starting point, contradicting part (b)'s claim of an everywhere absolutely continuous kernel.